Asked by Andre
Use the integral identity:
∫(a-1) (1/(1+x^2))dx=∫(1-1/a) (1/(1+u^2))du
for a>1 to show that:
arctan(a)+arctan(1/a)=π/2
∫(a-1) (1/(1+x^2))dx=∫(1-1/a) (1/(1+u^2))du
for a>1 to show that:
arctan(a)+arctan(1/a)=π/2
Answers
Answered by
Steve
after the integration, you have
arctan(1) - arctan(a) = arctan(1/a) - arctan(1)
π/4 - arctan(a) = arctan(1/a) - π/4
rearrange the terms and you're done.
arctan(1) - arctan(a) = arctan(1/a) - arctan(1)
π/4 - arctan(a) = arctan(1/a) - π/4
rearrange the terms and you're done.
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